CLASS 8 MATHS · PART 2 · CHAPTER 6 · STUDY NOTES
Algebra Play
Understand algebra play through short explanations, labelled models and original practice. Read one topic, explain it in your own words, then check.
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Letters reveal what the trick does
Replace the secret starting number with x and follow every instruction.Watch x disappear
- Variable
- A letter standing for a value that may vary.
Think of x, multiply by 3, add 15, divide by 3, then subtract x. The result is (3x + 15)/3 − x = 5.
The x terms cancel, so the final result does not depend on your starting number. Choosing examples checks the trick; simplifying the expression explains every case.
- 1
Write the starting value as x.
- 2
Translate each instruction into an expression.
- 3
Use brackets so division acts on all the required terms.
- 4
Simplify and see whether x remains.
Take a moment. What do you think?
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Question
How should 15 change if we want the final answer to be 7 instead?
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Encode the month and day using place value
Multiplying the month by 100 leaves two places for a valid day number.Separate the month and two-digit day
- Encoding
- Representing information by a number using a known rule.
An original date code is N = 100M + D + 57. Subtract 57 first. The last two digits then give day D, and the earlier digits give month M.
For N = 371, subtract 57 to get 314: month 3, day 14. Only valid calendar dates are allowed; a code does not make an impossible date valid.
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Question
Decode N = 1262 with this rule.
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The middle value reaches the top by two paths
Add neighbouring entries upward; subtract known neighbours when working downward.Two neighbours make the block above
- Number pyramid
- An arrangement where each entry is the sum of the two entries immediately below it.
For bottom row a,b,c, the next row is a + b and b + c, and the top is a + 2b + c.
For bottom 3,x,5 and top 20, write 3 + 2x + 5 = 20. This gives x = 6; the next row is 9 and 11.
- 1
Name the unknown bottom value x.
- 2
Use the weighted top expression a + 2b + c.
- 3
Solve the equation.
- 4
Fill the remaining blocks and check both routes to the top.
Let’s look a little closer.
Go deeper: More rows give new weights
With four bottom values a,b,c,d, repeated neighbouring sums give top a + 3b + 3c + d. These coefficients count how often each bottom entry contributes.
If the bottom entries are consecutive terms of a sequence where each term is the sum of the previous two, neighbouring sums move to later terms of that sequence. The pyramid rules explain the shift.
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Question
Find the top for bottom 4,7,2 without filling every block.
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A calendar grid has fixed offsets
In an ordinary seven-column calendar, the entry below a date is seven more.Right adds 1; down adds 7
- Grid relation
- A rule linking entries using their row and column positions.
A complete 2 by 2 block has entries a,a + 1,a + 7,a + 8. Their sum is 4a + 16.
If the sum is 60, then a = (60 − 16)/4 = 11, giving 11,12,18,19. The block must actually fit inside the month and cannot wrap across a row edge.
Go deeper: Solve shape-value grids with equations
If two identical symbols sum to 18, each is 9. Substitute this into another row equation to find a different symbol. A symbol keeps the same value wherever it occurs; different-looking symbols are not automatically required to have different values unless the puzzle says so.
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Question
A valid 2 by 2 block sums to 44. What is its top-left entry?
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Compare place-value expressions instead of guessing
Use each given nonzero digit once to make a two-digit number times a one-digit number.The tens contribution matches
- Systematic comparison
- Listing possibilities or proving an ordering so no allowed case is missed.
With digits 2,4,7, the best ordering for each multiplier gives 74×2 = 148, 72×4 = 288 and 42×7 = 294. The largest is 294.
For distinct positive digits p < q < r, compare (10q + p)r and (10r + p)q. Both contain 10qr; the difference is p(r − q) > 0.
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Question
What is the largest product using digits 1,5,8 once?
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Reverse the digits, then use powers of ten
Writing the digits in place-value form exposes a common factor.Digits have different place weights
- Two-digit place-value form
- A two-digit number with tens a and units b has value 10a + b.
The signed difference is (10a + b) − (10b + a) = 9(a − b), so it is divisible by 9. The nonnegative difference has quotient |a − b|.
The sum is 11(a + b), so it is divisible by 11. For example 62 − 26 = 36 and 62 + 26 = 88.
Go deeper: Repeat a three-digit block
If N is a three-digit block, repeating it gives 1000N + N = 1001N. Since 1001 = 7×11×13, dividing successively by 7,11,13 returns N.
Cycling digits a,b,c gives a sum of 111(a + b + c), which is divisible by both 3 and 37 because 111 = 3×37. Leading zeros must be handled as fixed-width blocks when they occur.
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Question
Why is 81 − 18 divisible by 9?
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Build an equation from what stays fixed
Name the unknown, write a relationship, then check the answer in the situation.Start with two wheels for every vehicle
- Equation
- A statement that two expressions are equal for the solution values.
A fictional car park has 12 vehicles, each with either 2 or 3 wheels, and 29 wheels total. If t have three wheels, the rest 12 − t have two.
3t + 2(12 − t) = 29 simplifies to t = 5. So there are 5 three-wheelers and 7 two-wheelers.
- 1
Choose an unknown and describe the other count using the total.
- 2
Write the total-wheel equation.
- 3
Distribute and collect like terms.
- 4
Check both vehicle count and wheel count.
Go deeper: Work backward in a repeated-operation puzzle
Suppose a fictional machine doubles tokens then takes 6. After three rounds it leaves zero. Reverse one round by adding 6 and dividing by 2: 0 → 3 → 4.5 → 5.25. If tokens must be whole, this setup is impossible; arithmetic solutions still need to satisfy the situation’s constraints.
For a pattern in fractions, compare whole sums, not individual terms. The sum of the first n odd numbers is n²; the next n odd numbers sum to (2n)² − n² = 3n², so their ratio is 1/3 for positive integer n.
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Question
A parent is five times a child’s age. In nine years the parent will be three times the child’s age. Find the child’s age now.
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